The problem is to factor the polynomial $x^3 - 7xy^2 - 6y^3$.

1. Problem Description

The problem is to factor the polynomial $x^3 - 7xy^2 - 6y^3$.

2. Solution Steps

We want to factor $x^3 - 7xy^2 - 6y^3$. We look for a factor of the form $(x - ay)$ where $a$ is a constant.

If $x = ay$ is a root, then $x - ay$ is a factor.

Substituting $x = ay$ into the equation $x^3 - 7xy^2 - 6y^3 = 0$, we have

$(ay)^3 - 7(ay)y^2 - 6y^3 = 0$

$a^3y^3 - 7ay^3 - 6y^3 = 0$

$y^3(a^3 - 7a - 6) = 0$

So we need to find the roots of the equation $a^3 - 7a - 6 = 0$.

By trial and error, we can test small integers.

If $a = -1$, $(-1)^3 - 7(-1) - 6 = -1 + 7 - 6 = 0$.

If $a = -2$, $(-2)^3 - 7(-2) - 6 = -8 + 14 - 6 = 0$.

If $a = 3$, $(3)^3 - 7(3) - 6 = 27 - 21 - 6 = 0$.

So $a = -1, -2, 3$ are the roots of $a^3 - 7a - 6 = 0$.

Thus, the polynomial factors as $(a + 1)(a + 2)(a - 3) = 0$.

This means $x + y$, $x + 2y$, and $x - 3y$ are factors of the expression $x^3 - 7xy^2 - 6y^3$.

Therefore, $x^3 - 7xy^2 - 6y^3 = (x+y)(x+2y)(x-3y)$.

We can expand $(x+y)(x+2y)(x-3y)$ to verify.

$(x+y)(x+2y) = x^2 + 2xy + xy + 2y^2 = x^2 + 3xy + 2y^2$

$(x^2 + 3xy + 2y^2)(x-3y) = x^3 - 3x^2y + 3x^2y - 9xy^2 + 2xy^2 - 6y^3 = x^3 - 7xy^2 - 6y^3$

3. Final Answer

$(x+y)(x+2y)(x-3y)$